Question

A tank of cross-sectional area A is filled with a liquid of specific weight γ1 as shown in Image (A). Show

that when a cylinder of specific weight γ2 and volume V is floated in the liquid as shown in Image (B), the

liquid level rises by an amount Δh = (

γ2

γ1

)( V

A

)

Answer 1

Consider that A is the cross-section area of the tank, $γ_1$ ​is the specific weight of the liquid in the tank,  $γ_2$ ​ is the specific weight of the cylinder immersed in the tank, and V is the total volume of the cylinder.

Let the initial height of water in the tank be $h_1$ ​and the initial volume of liquid be $V_1$. Therefore, the expression of the initial volume of the liquid is:

$V_1​=Ah_1​$

When the cylinder is submerged in the liquid of the tank, the liquid level rises to $h_2$. Therefore, the final volume of the tank is:

$V_2​=Ah_2​$

Now, the expression of the change in the volume of liquid of tank after cylinder submerged is: $ΔV=V_2​−V_1​$

This change in volume must be equal to that part of the volume of a cylinder that is fully or partially immersed in the liquid. Therefore, assume that X is the part of the volume of the cylinder which is submerged in liquid. Therefore, the above expression will become: $V_2​−V_1​=X$

Substitute $Ah_2​$ for $V_2$ ​and $Ah_1$ ​for $V_1$ :

$(Ah_2)-(Ah_1)=X$

$A(h_2-h_1)=X$

The condition for floating of the cylinder in the liquid is that the weight of the object must be equal to the buoyant force. Therefore, the weight of the cylinder is:

$W=mg........(1)$

Here, m is the mass of the cylinder and g is the acceleration due to gravity.

Let ${\rho _2}$ ​ be the density of water in the cylinder and V be the volume of a cylinder. Therefore, the expression of the density of the water cylinder is:

$density= \frac{volume}{mass}$

Substitute ${\rho _2}$ ​ for density, m for mass, and V for volume:

$\rho_2= \frac{v}{m} \implies m=\rho _2v$

Substitute ${\rho _2}V$  for m in equation (1)

$W=mg= \rho _2 Vg=(\rho _2 g) V=γ_2V$

Here, ${\gamma _2}$ is the specific weight of the cylinder.

Now, the expression of the buoyant force ${F_b}$  is:

$F _b =Vρ_1 g$

Here, ${F_b}$​ is a buoyant force, V is the volume of the cylinder that is submerged in the liquid, ${\rho _1}$ is the density of the liquid, and g is the acceleration due to gravity.

$Substitute X for V$

$F _b =Xρ_1 g =X(ρ_1 g )=X\gamma_1$

Here, ${\gamma _1}$ ​ is the specific weight of the liquid of the tank.

Now, write the expression for the condition of floating:

$W = {F_b}$

​Substitute ${\gamma _2}V$ for W and $X{\gamma _1}$ ​ for ${F_b}$ ​ :

$γ _2V=Xγ_1 \implies X= \frac{γ _2V}{γ _1}$

Substitute $A\left( {{h_2} - {h_1}} \right)$ for X

$A(h _2 −h _1 )=X$

$A(h _2−h _1 )=\frac{ γ _1}{γ _2} V$

$A\triangle h=\frac{ γ _1}{γ _2} V$

$\triangle h=\frac{ γ _1}{γ _2} \frac{ V}{A}$